Prove that${n \choose k} \frac{1}{n^k} \le 2^{1-k}$, concluding that: $2 \le e \le 3$ $${n \choose k} \frac{1}{n^k} \le 2^{1-k}$$ for all $1 \le k \le n$
by using $$e := lim_{n \rightarrow \infty} \Big(1+\frac{1}{1+n}\Big)^n$$
Show concluding that: $2 \le e \le 3$
My idea is the following:
${n\choose k}=\frac{n!}{k!(n-k)!}$ by using it :$${n \choose k} \frac{1}{n^k} \le 2^{1-k}=\frac{n!}{k!(n-k)!}\cdot\frac{1}{n^k} \le 2^{1-k}$$ I'm not sure that this step makes any sence since the exercise seems to be much more difficult now.
My second idea is that:
$${n\choose k}= {n\choose n-k}=\frac{n!}{{(n-k)}!(n-{(n-k)})!} = n!(n−k)!−k!$$ using the substitution:
$$n!(n−k)!-k!\frac{1}{n^k} \le  2^{1-k}$$ which doesn't makes the problem again more complicated.
I would appreciate your help a lot.
 A: The problem is equivalent to the following
$$\binom{n}{k}\frac{1}{n^k}\leq 2^{1-k}$$
$$\Rightarrow \binom{n}{k}\left( \frac{2}{n} \right)^k\leq 2$$
Now the term on the LHS are just the terms of the expansion of $(1+\frac{2}{n})^n$.Let $T_k$ denote the maximum term in the expansion.Our aim is to prove that $T_k\leq 2$
$$\frac{T_k}{T_{k-1}}\geq 1$$
$$\Rightarrow \frac{\frac{n!}{(n-k)!k!}\frac{2^k}{n^k}}{\frac{n!}{(k-1)!(n-k+1)!}\frac{2^{k-1}}{n^{k-1}}}\geq 1$$
This on simplification gives $k\leq 2\frac{(n+1)}{(n+2)}<2$.So k must be 0 or 1 .But we have $T_0=1$ and $T_1=n\times \frac{2}{n}=2$
Hence each term is less than 2 and our first part ,i.e, the inequality is proved.
Now for the second part consider
$$\binom{n}{k}\left( \frac{2}{n} \right)^k\leq 2$$
$$\Rightarrow \lim_{n\to \infty}\binom{n}{k}\left( \frac{2}{n} \right)^k\leq 2$$
$$\Rightarrow \frac{2^k}{k!}\leq 2$$
$$\Rightarrow \frac{1}{k!}\leq \frac{1}{2^{k-1}}$$
$$\Rightarrow 2\leq 1+\sum_{k=1}^{\infty}\frac{1}{k!}\leq 1+\sum_{k=1}^{\infty}\frac{1}{2^{k-1}}$$
$$\Rightarrow 2\leq e\leq 3$$
